Suppose that Young’s experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits. How far apart are the bright fringes near the center of the interference pattern?

Short Answer

Expert verified

The bright fringes near the center of the interference pattern are apart of 2.25 mm.

Step by step solution

01

Write the given data from the question

Blue-green light Wavelength, λ=500nm.

The slit separation, d=1.20mm.

The screen distance from the slit, D=5.4m.

02

Determine the formulas to calculate how far apart the bright fringes near the center of the interference pattern

The condition for the maxima in Young’s experiment is given as follows.

dsinθ=mλ …… (1)

Here, d is the distance between the slits, λis the wavelength, mis the order, andθis the angular separation.

03

Calculate how far apart the bright fringes near the center of the interference pattern

The maximum vertical distance from the center of the pattern is given by,

tanθsinθ=ymD

Substitute ymDfor sinθinto equation (1).

dymD=mλym=mλDd

Substitute 1 for m, 54 m for D, 500 nm for λand 1.20mmfor d into above equation.

ym=1×500×10-9×5.41.2×10-3=27×10-71.2×10-3=22.5×10-4m=2.25mm.

Hence the bright fringes near the center of interference pattern are apart of 2.25 mm.

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Most popular questions from this chapter

Find the slit separation of a double-slit arrangement that will produce interference fringes0.018radapart on a distant screen when the light has wavelengthλ=589nm.

Monochromatic green light, of wavelength 500 nm, illuminates two parallel narrow slits 7.70 mm apart. Calculate the angular deviation ( θin Fig. 35-10) of the third-order (m=3)bright fringe (a) in radians and (b) in degrees.

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(a) If we start at the indicated start point and travel along path 1, does the interference produce a maximum all along the path, a minimum all along the path, or alternating maxima and minima? Repeat for

(b) path 2 (along an axis through the sources) and

(c) path 3 (along a perpendicular to that axis).

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Transmission through thin layers. In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of r3and r4interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1,n2and n3the type of interference, the thin-layer thickness Lin nanometers, and the wavelength λin nanometers of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where L is missing, give the second least thickness or the third least thickness as indicated.

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